Volumes with Cross Sections: Triangles/Semicircles
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AP Calculus AB › Volumes with Cross Sections: Triangles/Semicircles
What integral gives the volume if the base is bounded by $y=1-x$ and $y=0$ on $0\le x\le 1$, with semicircular cross sections perpendicular to the $x$-axis?
$\displaystyle \int_{0}^{1} \frac{\pi}{4},(1-x)^2,dx$
$\displaystyle \int_{0}^{1} \pi,(1-x)^2,dx$
$\displaystyle \int_{0}^{1} \frac{\pi}{8},(1-x),dx$
$\displaystyle \int_{0}^{1} \frac{\pi}{8},(1-x)^2,dx$
$\displaystyle \int_{0}^{1} \frac{1}{2},(1-x)^2,dx$
Explanation
The skill here is to compute volumes of solids where cross-sections perpendicular to an axis are non-rectangular shapes like semicircles or equilateral triangles. The diameter of each semicircular cross-section is the distance between y=1-x and y=0, which is 1 - x. The area of a semicircle is (π/8) times the square of the diameter, so the cross-sectional area is (π/8)(1 - x)². Thus, the volume is given by integrating this area from x = 0 to x = 1. A tempting distractor is choice E, which uses π(1 - x)², treating the cross-sections as full circles instead of semicircles. In general, to find volumes with non-rectangular cross-sections, express the area of the shape in terms of the varying dimension and integrate along the axis of orientation.
What integral gives the volume if the base is bounded by $y=1$ and $y=\sin x$ on $0\le x\le \pi$, with equilateral triangular cross sections perpendicular to the $x$-axis?
$\displaystyle \int_{0}^{\pi} \frac{\sqrt{3}}{4},(1-\sin x),dx$
$\displaystyle \int_{0}^{\pi} \frac{\sqrt{3}}{4},(1-\sin x)^2,dx$
$\displaystyle \int_{0}^{\pi} \frac{1}{2},(1-\sin x)^2,dx$
$\displaystyle \int_{0}^{\pi} \frac{\pi}{8},(1-\sin x)^2,dx$
$\displaystyle \int_{0}^{\pi} \frac{\sqrt{3}}{4},(\sin x)^2,dx$
Explanation
This problem involves computing the volume of a solid with equilateral triangular cross-sections perpendicular to the x-axis, a key skill in understanding non-rectangular cross-sections. The side length of each triangle is the vertical distance between the curves y=1 and y=sin x, which is 1 - sin x. The area of an equilateral triangle is (√3/4) s², where s = 1 - sin x. Therefore, the area simplifies to (√3/4) (1 - sin x)². The volume is the integral of this area from x=0 to x=π. A common mistake is to use (√3/4) (1 - sin x), as in choice B, which forgets to square the side length. Remember, for any cross-section shape, determine the base length, apply the area formula correctly, and integrate along the perpendicular axis.
What integral gives the volume if the base is bounded by $y=4-x$ and $y=x$ for $0\le x\le 2$, with semicircular cross sections perpendicular to the $x$-axis?
$\displaystyle \int_{0}^{2} \frac{\pi}{8},(4-2x),dx$
$\displaystyle \int_{0}^{2} \pi,(4-2x)^2,dx$
$\displaystyle \int_{0}^{2} \frac{\pi}{8},(4-2x)^2,dx$
$\displaystyle \int_{0}^{2} \frac{\pi}{4},(4-2x)^2,dx$
$\displaystyle \int_{0}^{2} \frac{1}{2},(4-2x)^2,dx$
Explanation
This problem involves computing the volume of a solid with semicircular cross-sections perpendicular to the x-axis, a key skill in understanding non-rectangular cross-sections. The diameter of each semicircle is the vertical distance between the curves y=4-x and y=x, which is 4-2x. The area of a semicircle is (1/2) π r², where r = (4-2x)/2. Therefore, the area simplifies to π (4-2x)² / 8. The volume is the integral of this area from x=0 to x=2. A common mistake is to use π (4-2x) / 8, as in choice B, which forgets to square the diameter in the formula. Remember, for any cross-section shape, determine the base length, apply the area formula correctly, and integrate along the perpendicular axis.
What integral gives the volume if the base is bounded by $x=0$ and $x=3y$ for $-1\le y\le 2$, with equilateral triangular cross sections perpendicular to the $y$-axis?
$\displaystyle \int_{-1}^{2} \frac{\pi}{8},(3y)^2,dy$
$\displaystyle \int_{-1}^{2} \frac{\sqrt{3}}{4},(3y)^2,dy$
$\displaystyle \int_{-1}^{2} \frac{\sqrt{3}}{4},(3y),dy$
$\displaystyle \int_{-1}^{2} \frac{1}{2},(3y)^2,dy$
$\displaystyle \int_{-1}^{2} \frac{\sqrt{3}}{4},(y)^2,dy$
Explanation
This problem involves computing the volume of a solid with equilateral triangular cross-sections perpendicular to the y-axis, a key skill in understanding non-rectangular cross-sections. The side length of each triangle is the horizontal distance between the curves x=0 and x=3y, which is 3y (noting y from -1 to 2, but area squares it). The area of an equilateral triangle is (√3/4) s², where s = 3y. Therefore, the area simplifies to (√3/4) (3y)². The volume is the integral of this area from y=-1 to y=2. A common mistake is to use (√3/4) (3y), as in choice B, which forgets to square the side length. Remember, for any cross-section shape, determine the base length, apply the area formula correctly, and integrate along the perpendicular axis.
What integral gives the volume if the base is bounded by $y=\sqrt{x}$ and $y=0$ on $0,4$, with equilateral triangular cross sections perpendicular to the $x$-axis?
$\displaystyle \int_{0}^{4} \frac{1}{2},(\sqrt{x})^2,dx$
$\displaystyle \int_{0}^{4} \frac{\sqrt{3}}{4},(\sqrt{x}),dx$
$\displaystyle \int_{0}^{4} \frac{\sqrt{3}}{4},(\sqrt{x})^2,dx$
$\displaystyle \int_{0}^{4} \frac{\pi}{4},(\sqrt{x})^2,dx$
$\displaystyle \int_{0}^{4} \frac{\sqrt{3}}{4},x,dx$
Explanation
This problem involves finding the volume when cross-sections are equilateral triangles perpendicular to the x-axis. The base region is bounded by y = √x and y = 0, so at each x-value, the side length of the equilateral triangle equals √x. For an equilateral triangle with side length s, the area is A = (√3/4)s². Therefore, A(x) = (√3/4)(√x)² = (√3/4)x. The volume integral is ∫_${0}^{4}$ (√3/4)x dx, which matches option B since (√x)² = x. Option C incorrectly uses √x instead of (√x)² in the integrand, failing to square the side length. Remember: for any cross-section shape, express its area in terms of the base dimension, then integrate.
What integral gives the volume if the base is bounded by $y=\sin x$ and $y=0$ on $0\le x\le \pi$, with semicircular cross sections perpendicular to the $x$-axis?
$\displaystyle \int_{0}^{\pi} \frac{1}{2},(\sin x)^2,dx$
$\displaystyle \int_{0}^{\pi} \frac{\pi}{8},(\sin x)^2,dx$
$\displaystyle \int_{0}^{\pi} \pi,(\sin x)^2,dx$
$\displaystyle \int_{0}^{\pi} \frac{\pi}{8},\sin x,dx$
$\displaystyle \int_{0}^{\pi} \frac{\pi}{4},(\sin x)^2,dx$
Explanation
The skill here is to compute volumes of solids where cross-sections perpendicular to an axis are non-rectangular shapes like semicircles or equilateral triangles. The diameter of each semicircular cross-section is the distance between y=sin x and y=0, which is sin x. The area of a semicircle is (π/8) times the square of the diameter, so the cross-sectional area is (π/8)(sin x)². Thus, the volume is given by integrating this area from x = 0 to x = π. A tempting distractor is choice E, which uses π(sin x)², treating the cross-sections as full circles instead of semicircles. In general, to find volumes with non-rectangular cross-sections, express the area of the shape in terms of the varying dimension and integrate along the axis of orientation.
What integral gives the volume if the base is bounded by $y=2x$ and $y=x^2$ for $0\le x\le 2$, with semicircular cross sections perpendicular to the $x$-axis?
$\displaystyle \int_{0}^{2} \frac{\pi}{8},(2x-x^2),dx$
$\displaystyle \int_{0}^{2} \frac{1}{2},(2x-x^2)^2,dx$
$\displaystyle \int_{0}^{2} \frac{\pi}{8},(2x-x^2)^2,dx$
$\displaystyle \int_{0}^{2} \pi,(2x-x^2)^2,dx$
$\displaystyle \int_{0}^{2} \frac{\pi}{4},(2x-x^2)^2,dx$
Explanation
The skill here is to compute volumes of solids where cross-sections perpendicular to an axis are non-rectangular shapes like semicircles or equilateral triangles. The diameter of each semicircular cross-section is the distance between y=2x and y=x², which is 2x - x². The area of a semicircle is (π/8) times the square of the diameter, so the cross-sectional area is (π/8)(2x - x²)². Thus, the volume is given by integrating this area from x = 0 to x = 2. A tempting distractor is choice C, which uses π(2x - x²)², treating the cross-sections as full circles instead of semicircles. In general, to find volumes with non-rectangular cross-sections, express the area of the shape in terms of the varying dimension and integrate along the axis of orientation.
What integral gives the volume if the base is bounded by $x=y^2$ and $x=4$ with semicircular cross sections perpendicular to the $y$-axis?
$\displaystyle \int_{-2}^{2} \frac{1}{2},(4-y^2)^2,dy$
$\displaystyle \int_{0}^{4} \frac{\pi}{8},(4-x^2)^2,dx$
$\displaystyle \int_{-2}^{2} \pi,(4-y^2)^2,dy$
$\displaystyle \int_{-2}^{2} \frac{\pi}{8},(4-y^2),dy$
$\displaystyle \int_{-2}^{2} \frac{\pi}{8},(4-y^2)^2,dy$
Explanation
The skill here is to compute volumes of solids where cross-sections perpendicular to an axis are non-rectangular shapes like semicircles or equilateral triangles. The diameter of each semicircular cross-section is the distance between x=4 and x=y², which is 4 - y². The area of a semicircle is (π/8) times the square of the diameter, so the cross-sectional area is (π/8)(4 - y²)². Thus, the volume is given by integrating this area from y = -2 to y = 2. A tempting distractor is choice E, which uses π(4 - y²)², treating the cross-sections as full circles instead of semicircles. In general, to find volumes with non-rectangular cross-sections, express the area of the shape in terms of the varying dimension and integrate along the axis of orientation.
What integral gives the volume if the base is bounded by $y=x$ and $y=x^3$ for $0\le x\le 1$, with semicircular cross sections perpendicular to the $x$-axis?
$\displaystyle \int_{0}^{1} \frac{1}{2},(x-x^3)^2,dx$
$\displaystyle \int_{0}^{1} \pi,(x-x^3)^2,dx$
$\displaystyle \int_{0}^{1} \frac{\pi}{4},(x-x^3)^2,dx$
$\displaystyle \int_{0}^{1} \frac{\pi}{8},(x-x^3)^2,dx$
$\displaystyle \int_{0}^{1} \frac{\pi}{8},(x-x^3),dx$
Explanation
The skill here is to compute volumes of solids where cross-sections perpendicular to an axis are non-rectangular shapes like semicircles or equilateral triangles. The diameter of each semicircular cross-section is the distance between y=x and y=x³, which is x - x³. The area of a semicircle is (π/8) times the square of the diameter, so the cross-sectional area is (π/8)(x - x³)². Thus, the volume is given by integrating this area from x = 0 to x = 1. A tempting distractor is choice C, which uses π(x - x³)², treating the cross-sections as full circles instead of semicircles. In general, to find volumes with non-rectangular cross-sections, express the area of the shape in terms of the varying dimension and integrate along the axis of orientation.
What integral gives the volume if the base is bounded by $y=\tan x$ and $y=0$ on $0\le x\le \frac{\pi}{4}$, with semicircular cross sections perpendicular to the $x$-axis?
$\displaystyle \int_{0}^{\pi/4} \frac{\pi}{8},\tan x,dx$
$\displaystyle \int_{0}^{\pi/4} \pi,(\tan x)^2,dx$
$\displaystyle \int_{0}^{\pi/4} \frac{1}{2},(\tan x)^2,dx$
$\displaystyle \int_{0}^{\pi/4} \frac{\pi}{8},(\tan x)^2,dx$
$\displaystyle \int_{0}^{\pi/4} \frac{\pi}{4},(\tan x)^2,dx$
Explanation
This problem involves computing the volume of a solid with semicircular cross-sections perpendicular to the x-axis, a key skill in understanding non-rectangular cross-sections. The diameter of each semicircle is the vertical distance between the curves y=tan x and y=0, which is tan x. The area of a semicircle is (1/2) π r², where r = (tan x)/2. Therefore, the area simplifies to π (tan x)² / 8. The volume is the integral of this area from x=0 to x=π/4. A common mistake is to use π (tan x)² / 4, as in choice C, which would be for full circles with radius (tan x)/2. Remember, for any cross-section shape, determine the base length, apply the area formula correctly, and integrate along the perpendicular axis.